Are you specifically looking for a "direct", self-contained proof? I guess you can show this by following the proof of Stokes with a bit more attention, but in practice taking the smooth Stokes for granted and applying a quick mollifier-argument would probably be far more efficient. Sadly I have no good reference for either.

I mean the second quote in the question, which was talking about domains of "class $C^1$ or Lipschitz". And you are right that without the comma the sentence can only be read in one way, but that could have been a typo, so maybe we have to wait for clarification.

It is the image of a Lipschitz function, but not the graph; in general one uses Lipschitz-domains precisely to avoid the type of cusp you have in your example, as they tend to cause all kind of problems in PDE. But you are right, it is not clear if the question meant (Lipschitz connected) or Lipschitz, connected domains. However I have never seen the former anywhere, while the latter is standard and also the one used in the second example.

That is not a Lipschitz-domain though, as no matter how you rotate it, the boundary is not the graph of a Lipschitz-function in any neighborhood of the origin.

The second condition is actually kind of a complement of the first. (A) effectively says that the density at the boundary is bounded away from 1 and the other that it is bounded away from 0. If you replace $\Omega$ with its complement, you turn one into the other. And both hold rather obviously for Lipschitz domains, as can be shown by looking at a local graph.

I don't understand your notation. If $f$ is only defined on the boundary of the unit cube, how can you have $u=f$ on $\mathbb{R}^d$? In any case, the first approach to such a problem would be to take the Fourier transform with respect to $x$, possibly even starting in 1d for simplicity. This gives you a bunch of independent ODEs which can be explicitly solved and will tell you a lot about the behaviour and about what initial and boundary conditions could work.

I think the only problem with the boundary would be if you accidentally take out a whole set of positive $n-1$-dimensional measure at once. E.g. if your set is the ball $B_1(0)$ and you take out $B_s(0)$ for a parameter $s\in [0,1]$. Then this fails to be continuous at $s=1$ as original and removed boundary line up. But taking any other center for the ball works. And this should generalize, because this approach failing for every family of removed sets should imply that the boundary is not rectifiable.

How do you define the flat norm on Cacciopoli sets? As the norm of the current corresponding to the integration of its boundary? If so then yes, you can simply take out larger and larger chunks of the set until it vanishes, thereby path-connecting it to the empty-set. It gets a bit more interesting though, if you modify the flat norm to not allow extra boundaries, then you get the homology-classes of the manifold instead.

I don't have the book to hand, but the key to coercivity normally is the boundedness from below. Folland's condition implies that the real part of $D(u,u)$ can become a little negative, but not too much. In particular there exists an infimum and any minimizing sequence will be bounded. With your condition that is not the case, as $D(u,u)$ could simply go to $-\infty$.

You cannot enforce continuity without controlling the derivative. You can approximate that discontinuous solution by continuous solutions deviating from the optimum only in $(T-\epsilon,T)$, which for $\epsilon \to 0$ will converge pointwise and in value to that jump. So assuming the minima of $F(.,T)$ and $B$ do not accidentally sync up, the problem has no continuous minimizer.

Since $u\in L^p((0,T);L^p(\Omega))$, you can find a sequence of simple functions $v_n: (0,T) \to L^p(\Omega)$ and similar simple functions $w_n: (0,T) \to L^p(\Omega)$ for $\nabla u$. If I have it correctly, the idea should now be to modify $v_n$ by taking a set on which both $v_n$ and $w_n$ are constant and replacing the value there with one of $u$ that is close. You might need some Egorov and diagonal sequences to make it work properly though.

Controlling curvature avoids that problem, assuming you also upgrade piecewise smooth to smooth. But there is also multiplicity. Take $X_n$ to be two different disjoint manifolds both converging to same limit, then the Hausdorff limit is just one copy of the limit. You can even do the same with a connected manifold by taking the boundary of a Möbius-strip of width $1/n$, which converges to a single circle.

Take any manifold $Y$ and approximate it by a sawtooth of higher and higher frequency but lower and lower height. This way you can even get $H^k(X_n) \to \infty$ and still have convergence in the Hausdorff distance.

I am not disagreeing with your findings, that algebraic geometry is probably overrepresented, but I am not sure how significant looking at the top 3 areas is without normalizing them in some way (e.g. I am not surprised to see PDE in that list, as it is one of the largest fields) and without giving discussing the size of the tail (e.g. are most of the articles in the top 3, or are there just slightly more than for place 4 and so on).

@l'étudiant Then this trick does not work anymore. You could maybe still use it to analyse the operators $\partial^2_x $ and $\partial^2_y $ on $\Omega$ and then do some semi-group theory regarding $(\partial_y^2)^{-1} \partial_x^2$, but I am not entirely sure if your new problem is even well posed.